# What's the ratio?

1. May 4, 2007

### MeJennifer

Observer A and B and a far away star are at rest with respect to each other, A and B are removed r meters from each other. B instantly starts to circumnavigate A and is making sure he is keeping a constant distance r from A. After one complete circumnavigation (he is using the star as a reference) he is looking at his odometer.

Does his meter show $2\pi r$?

Last edited: May 4, 2007
2. May 4, 2007

### cesiumfrog

..if he's far away from any star/mass, then isn't the answer yes? At least assuming the angular velocity is low? But how do you implement an odometer in empty space?

3. May 7, 2007

### paw

Well the presence of a distant star would allow A to navigate to a high degree of precision. If A and B's masses are small there should be no classical correction for gravity. If A's velocity is small there should be no SR corrections. If there are no nearby significant masses there should be no GR corrections.

And I assume A and B are large enough that there'd be no quantum corrections necessary.

I'd have to say yes, 2 pi r is the odometer reading. But then I'm only a chemist........

4. May 7, 2007

### JesseM

In the case of a "static solution" like the Schwarzschild spacetime, can't you do something like suspending a tape measure along the route the ship takes, and mark it in such a way that in the local inertial rest frame of a freefalling observer passing next to a given small section and instantaneously at rest with respect to it, the marks on that section are the correct distance apart?

Last edited: May 7, 2007
5. May 7, 2007

### pervect

Staff Emeritus
In the limit, as r->0, he will read 2 \pi r, but for large r he may read something different.

The plane through A may have some non-zero Gaussian curvature.

The formula in Wiki http://en.wikipedia.org/wiki/Curvature
(I haven't double checked this with my textbooks) gives

$$K = \lim_{r \rightarrow 0} (2 \pi r - \mbox{C}(r)) \cdot \frac{3}{\pi r^3}$$

The existence of this limit implies that

$$\lim_{r \rightarrow 0} (2 \pi r - \mbox{C}(r))} = 0$$

Last edited: May 7, 2007
6. May 7, 2007

### country boy

If the circumnavigation is an orbit, and
1) if A is much more massive than B, then the odometer will read 2*pi*r.
2) if A and B are the same mass, the odometer will read pi*r.
3) if A is much less massive than B, the odometer will read zero.